A jet of water leaves the tube at an angle of 30 degrees to the horizon at a speed

A jet of water leaves the tube at an angle of 30 degrees to the horizon at a speed of 10 m / s. Find the flight range time as well as the highest altitude.

Given:

v = 10 meters per second – the speed of the water stream;

a = 30 degrees – the angle of inclination of the water jet to the horizon;

g = 10 meters per second squared – the acceleration of gravity (constant near the surface of the earth).

It is required to determine L (meter) – the range of the water jet and the maximum flight altitude H (meter).

Let’s find the time of the flight of the water jet to the maximum height:

t = v * sin (a) / g = 10 * sin (30) / 10 = 1 * sin (30) = 1 * 0.5 = 0.5 seconds.

Then the entire jet will be in flight:

t1 = 2 * t = 2 * 0.5 = 1 second.

The flight range will be equal to:

L = v * cos (a) * t1 = 10 * cos (30) * 1 = 10 * 0.87 = 8.7 meters.

The maximum height will be:

H = v * sin (a) * t – g * t2 / 2 = 10 * 0.5 * 0.5 – 10 * 0.52 / 2 = 2.5 – 1.25 = 1.25 meters.

Answer: the range will be 8.7 meters, the height – 1.25 meters.